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Proof of non-existence of perfect cuboids

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Let a, b, c, d, e, f and g be positive integers. We consider that a, b and c have no common prime factors and a, b and c are the edge lengths, d, e and f are the diagonal lengths and g is the space diagonal length. If a perfect cuboid exists, the following equations hold.
a2+b2=d2
b2+c2=e2
c2+a2=f2
a2+b2+c2=g2
a is odd and b and c are even because a, b and c are not all even and if two edges are odd, then the diagonal is not a square number. And d and f are odd and e is even. Let i, ki, mi and ni be positive inetgers and 1i6 hods. We consider that mi and ni are relatively prime and the odd/even number of mi and ni are different. By the form of Pythagorean triple, the following equations hold.
a=(m12n12)k1=(m22n22)k2=(m52n52)k5
b=2k2m2n2=(m32n32)k3=2k6m6n6
c=2k3m3n3=2k1m1n1=2k4m4n4
d=(m22+n22)k2=(m42n42)k4
e=(m32+n32)k3=2k5m5n5
f=(m12+n12)k1=(m62n62)k6
g=(m42+n42)k4=(m52+n52)k5=(m62+n62)k6 ...(1)
k3 is even and the others ki are all odd. We consider that k4, k5 and k6 are mutually prime to each other since if two of the diagonal lengths have a common prime factor, then the length of the space diagonal and the length of the other diagonal will also have the prime factor, three edge lengths will have it too and this contradicts the condition that a, b and c have no common prime factors. Let r be an odd positive integer. Since g has factors k4, k5 and k6,
g2=((m52n52)k5)2+(2k6m6n6)2+(2k4m4n4)2=(rk4k5k6)2 ...(2)
holds. Let s and u be odd positive integers and t be an even positive integer. By this equation, the following equations hold.
((m52n52)k5)2+(2k6m6n6)2=sk42
(2k6m6n6)2+(2k4m4n4)2=tk52
(2k4m4n4)2+((m52n52)k5)2=uk62
By these equations,
sk42+tk52+uk62=2(k4k5k6)2
holds. Let v and w be odd positive integers and x be an even positive integer. By this equation, the following equations hold.
sk42+tk52=vk62
tk52+uk62=wk42
uk62+sk42=xk52
By these equations,
wk42+xk52+vk62=(2rk4k5k6)2 ...(3)
holds. Comparing the equation (3) with the equation (2), we find that the following equations hold since the equations (2) and (3) must hold for any k4, k5 and k6 satisfying the equation (1).
v=(4m6n6)2
w=(4m4n4)2
x=(2(m52n52))2
It becomes a contradiction since v and w are odd. From the above, there are no perfect cuboids. (Q.E.D.)

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